as-500/hw18/text.motes

Homework 18, due Nov 1
The N II spectrum forms from N + , which is iso-electronic with O III / O 2+. This means that they have the same number of orbiting electrons although the nuclear charge is different. The structure of the energy levels of N II will be similar to O III although the energies and As will bedifferent. In homework 15 did O III. For this we will do N II. Go to NIST and look at the energy levels given there. 1 The lowest terms of O III are 3 P, 1 D, and 1 S, as shown in Lecture 15, top left slide of page 3. The energies of the levels are also given on that slide. How does the energies of the levels in N II compare with O III? What is the net nuclear charge of an orbiting valence electron in N II? What is the net nuclear charge for O III? For both, the net charge is the nuclear charge minus the number of inner electrons that screen the nuclear charge.
    
    The energy levels have the same ratios, so the absolute value of each level for one ion is a scalar multiple of the absolute value of the energy of the level of the other ion.

    For N II, Z[N II] = 7. For O III, Z[O III] = 8.

    Net nuclear charge Q[N II] = 7 - 1 = 6, and Q[O III] = 8 - 2 = 6.

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2. Use these energies to write out the Boltzmann factor, exp −E/kT, for the 1 D and 1 S levels.

    @ 10000 K, kT = 0.6333621 Rydberg
    Boltzmann Factor β = exp(-E/kt) = exp(-E/(0.6333621 R))

            1D         1S
    E       1.396e-1   2.979e-1
    β       4.0111     6.2480e-1

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3. For O III the 1 S – 1 D transition has a wavelength of 4363A. The wavelength of the 1 D – 3 P 2 level is 5007A. Use NIST and air wavelengths to find the wavelength of these two transitions for N II.

    E = hf = hc/λ, λ = hc/E, where E = E₁ - E₂ (a transition from higher energy E₂  to lower energy E₁).
    
    N II         E1 (R)        E2 (R)       wavelength (A)
    1S → 1D     2.9788E-01    1.3957E-01    5755.9878991494
    1D → 3P2    1.3957E-01    1.1920E-03    6585.0757723516

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4. Use NIST to find the total Einstein A for the 1 S – 1 D and 1 D – 3 P 2 lines. How do they compare with the As for O III?

    Just looking them up in the table by wavelength,
    
    N II       wavelength (A)    A (1/s)
    1S → 1D   5755.9878991494   1.14 
    1D → 3P2  6585.0757723516   8.65e-06
    
    By the way, I see that my wavelengths are off a bit. Maybe difference between air and vacuum?

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5. What is the Boltzmann factor that predicts the LTE population ratio of the 1 D and 1 S levels? Assume LTE to predict the ratio of the 5755 to 6584 lines.

    I realize now as I inspect homework 15 more closely that I should be doing these things as a function of T. Alright,

    exp(-E₁₂/kT) =  exp(-(E₁ᴅ - E₁ꜱ)/kT)

    exp(((E₁ᴅ - E₁ꜱ)/k)/T) 
        = exp((0.1384 R - 0.1583)/(6.334e-6 R K⁻¹)/T) 
        = exp((-3147 K)/T).

    Assuming LTE, the line ratio is given by g₁/g₂ A₁/A₂ λ₂/λ₁ exp((-3147 K)/T).

    
    g₁/g₂ A₁/A₂ λ₂/λ₁ = 3.316e-5.

    Seems small just based on that we'd be studying something in class that has a lot of action, but I think this should be expected considering the disparity in the einstein probabilities. I guess this just slides the action up or down the temperature scale, anyway.

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6. The ratio observed in the Orion Nebula is 0.02 ±0.003. What is the kinetic temperature, including the uncertainty? (This assumes LTE – that is not correct for the Orion Nebula.) 
    
    3.316e-5 exp((3147 K)/T) = 0.02 ±0.003.

           Note: I've gone back and forth about this, because I would think this exp() factor should be negative, but from 

                g₁/g₂ nₜₒₜ/U U/nₜₒₜ exp(-(E₁ - E₀))/exp(-(E₂ - E₀)),

           I get g₁/g₂ exp(E₂ - E₁).

    3.316e-5 exp((3147 K)/T) = 0.02 ±0.003.

    ln(3.316e-5) (3147 K)/T = ln(0.02 ±0.003).

    T = ln(3.316e-5)/ln(0.02 ±0.003) (3147 K) = 7.966e3 K 
        ≈ 8000 K. 

    This is quite high, like the photosphere of a star a bit hotter than Sol. I expected to get something more like 4000K, because of a previous assignment. I'm thinking there's an error here, but not an egregious one.

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We have demonstrated that Mathmatica cannot solve HW 17. Redo this with a spreadsheet. First derive and show the equations that will go in each column of the spreadsheet. In the spreadsheet make the first columns a log10 range in z with 10 < z < 1e5. Makecolumns giving the kinetic temperature, total hydrogen density, and age of the universe. Next use the Saha equation and electron conservation to write down the expression for the H + and H 0 density in terms of the temperature, by solving the quadratic.

    Approximate relationship of redshift z to thermodynamic properties:

        ◆ T kinetic = 2.7 (1+z) K
        ◆ n(H) ≈ 2.51×10⁻⁷ (1+z)³ cm ⁻³
        ◆ Age ≈ 1.3×10¹⁰ (1+z)\^-1.5  yr

    The Saha equation for ionized hydrogen is
        nᵢₒₙ nₑ/nₐₜₒₘ = Uᵢₒₙgₑ/Uₐₜₒₘ (2πmₑkT/h²)^(3/2) exp(-Eᵢₒₙ/kT).

    Checking units:

        2πmₑkT/h² → g R (R*s)⁻² 
                  → g R⁻¹ s⁻² 
                  → g (g cm² s⁻²)⁻¹ s⁻²
                  → cm⁻²

        (2πmₑkT/h²)^3/2 → cm⁻³

    Of course, nᵢₒₙ = nₑ, for ionized hydrogen.

    (2πmₑkT/h²)^(3/2) = 2.415e15 cm⁻³ K^-(3/2) T^(3/2).

    Established in previous assignments.

        Eᵢₒₙ = 1 R.
        gₑ = 2.
        Uᵢₒₙ = 1.
        Uₐₜₒₘ = 2.

    So,
        nᵢₒₙ²/nₐₜₒₘ = (2.415e15 cm⁻³ K^-(3/2) T^(3/2)) exp(-(1 R)/kT).

    Set
        α(T) = 2.415e15 cm⁻³ K^-(3/2) T^(3/2), and
        β(T) = exp(-(1 R)/kT) = exp(-157900 K T⁻¹), and
        γ(T) = α β.

    Then, nᵢₒₙ²/nₐₜₒₘ = γ.

    Particle conservation gives nₜₒₜ - nᵢₒₙ = nₐₜₒₘ.

    To find an expression for the ionized hydrogen fraction, substitute for

        nᵢₒₙ²/(nₜₒₜ - nᵢₒₙ) = γ.

    nᵢₒₙ² + γ nᵢₒₙ - γ nₜₒₜ = 0.

    Quadratic formula gives
    
        nᵢₒₙ(nₜₒₜ,T) = 1/2 (-√γ √(γ + 4nₜₒₜ) - γ) and

        nᵢₒₙ(nₜₒₜ,T) = 1/2 (√γ √(γ + 4nₜₒₜ) - γ).

    For neutral hydrogen fraction,

        (nₜₒₜ - nₐₜₒₘ)²/nₐₜₒₘ = γ.

    (nₜₒₜ - nₐₜₒₘ)(nₜₒₜ - nₐₜₒₘ) = γ nₐₜₒₘ.

    nₜₒₜ² - 2 nₐₜₒₘ nₜₒₜ + nₐₜₒₘ² = γ nₐₜₒₘ.

    nₐₜₒₘ² + nₜₒₜ² - 2 nₐₜₒₘ nₜₒₜ - γ nₐₜₒₘ = 0.

    nₐₜₒₘ² - nₐₜₒₘ(2 nₜₒₜ + γ) + nₜₒₜ² = 0.

    Quadratic formula gives

        nₐₜₒₘ(nₜₒₜ,T) = 1/2 (± √γ √(γ + 4 nₜₒₜ) + γ + 2 nₜₒₜ).

    Reminder: α(T) = 2.415e15 cm⁻³ K^-(3/2) T^(3/2), and
              β(T) = exp(-157900 K⁻¹ T).

So, I have
    
    nᵢₒₙ(nₜₒₜ,T) = 1/2 (√γ √(γ + 4nₜₒₜ) - γ), and

    nₐₜₒₘ(nₜₒₜ,T) = 1/2 (√γ √(γ + 4 nₜₒₜ) + γ + 2 nₜₒₜ), with
    
    γ(T) = 2.415e15 cm⁻³ K^-(3/2) T^(3/2) exp(-157900 K T⁻¹).

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Make columns giving the H + and H 0 density at each redshift. Do not take the difference to two nearly identical numbers! Use different solutions for the quadratic to get the H + and H 0 densities. Plot each over the z range 10 to 1e5. The universe recombined at roughly z≈1000. 

    OK. I was stuck on this for a long time. I used the above equations, but I have what seems like numerical instability at high redshift. Perhaps I've used an unstable expression, but I think what I'm seeing here is a problem with librecalc when it computes √γ √(γ + 4nₜₒₜ) - γ. It seems that it's leaving out the nₜₒₜ factor where γ is very large, because of the disparity in order, I suppose. This is unfortunate, but I can't see how to fix it at the moment. However, the graph at least shows the general characteristics I'm looking for, with the ionization front at z≃1000. Plot attached.

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Find the brems opacity, recalling that the units are cm -1 . Make columns giving the photon mean free path (the inverse of the opacity) and the visible diameter of the universe. At what z do they cross? What is the temperature at that redshift? That is the photosphere we observe. Use the Wien displacement theorem to find the temperature of the blackbody we observe today.


Plot axes must be labeled with a font that can be read on a screen. Do not use a tiny font. Clearly label all lines on the plot. Especially clear and well-done examples may be posted as the solution. There will be extra credit if I use yours.